A solution prepared by mixing 10 mL of a 0.10 M solution of the R enantiomer of a compound and 30 mL of a 0.10 M solution of the S enantiomer was found to have an observed specific rotation of +4.8. W

Question

A solution prepared by mixing 10 mL of a 0.10 M solution of the R enantiomer of a compound and 30 mL of a 0.10 M solution of the S enantiomer was found to have an observed specific rotation of +4.8. What is the specific rotation of each of the enantiomers?

TestHub · Accepted Answer

Ans. (A) [α]R = -9.6, [α]S = +9.6.

Sol. 10 mL x 0.10 M = 1 mmol of R and 30 mL x 0.10 M = 3 mmol of S. 1 mmol of R with 1 mmol of S forms a racemic 2 mmol, leaving 2 mmol of S in excess out of 4 mmol total: enantiomeric excess = 50% S. ee = (observed rotation / rotation of pure enantiomer) x 100%, so 50 = (+4.8 / x) x 100 and x = +9.6 for the S enantiomer (the one in excess); the R enantiomer has the opposite value, -9.6.

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